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Licemer1 [7]
3 years ago
13

A hazard sign has 3 identical

Physics
1 answer:
melomori [17]3 years ago
5 0

Answer and Explanation: To know how much tape he will need, we have to calculate the perimeter of each parallelogram-shaped stripe.

Perimeter is the sum of all the sides of a figure.

For a parallelogram:

P = 2*length + 2*width

So, we need to determine width and length of the stripe.

Width is 3 inches. Length is the hypotenuse of the right triangle, whose sides are 6 and 18 inches. Then, length is

h=\sqrt{18^{2}+6^{2}}

h=\sqrt{360}

h = 19 in

Perimeter of the first stripe is

P = (2*19) + (2*3)

P = 44 inches

The hazard sign has 3 stripes. So total perimeter is

P_{t}= 44 + 44 + 44

P_{t}= 132 inches

To outline the parallelogram-shaped stripes, Charles need a total of 132 inches of tape. Since one roll has 144 inches, he will have enough tape to finish the job.

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Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is
kumpel [21]

Answer:

7800kg/m³

Explanation:

Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is

Given the density of iron = 7.8 g/cm3.

The SI units must be in kg/m³

7.8g = 7.8/1000 kg

7.8g = 0.0078kg

1cm³ = 0.000001m³

7.8g/cm³

= 0.0078/0.000001 kg/m³

= 7800kg/m³

Hence the density in SI unit is 7800kg/m³

4 0
3 years ago
All contraceptives prevent STIs true or false
NNADVOKAT [17]
False. Not all contraceptives prevent STIs.

Contraceptives are items used or taken to prevent conception. These are birth control pill, patch, ring, IUD, birth control shots, and condoms. Among these contraceptives, condoms are the only type of birth control that prevents STIs or Sexually Transmitted Infections.
7 0
3 years ago
A box is pulled with a horizontal force of 500N and moves 5m what is work done
dalvyx [7]
The answer to the question is shown below:

We all know that formula for solving work done is the force multiplied by distance covered:
Work done = Force x distance
Distance = 5m
Force = 500 N
Work done = 500 N * 5m
Work done = 2500 J

4 0
3 years ago
An assault rifle fires an eight-shot burst in 0.40 s. Each bullet has a mass of 7.5 g and a speed of 300 m/s as it leaves the gu
myrzilka [38]

Answer:

The average recoil force on the gun during that 0.40 s burst is 45 N.

Explanation:

Mass of each bullet, m = 7.5 g = 0.0075 kg

Speed of the bullet, v = 300 m/s

Time, t = 0.4 s

The change in momentum of an object is equal to impulse delivered. So,

F\times t=mv\\\\F=\dfrac{mv}{t}

For 8 shot burst, average recoil force on the gun is :

F=\dfrac{8mv}{t}\\\\F=\dfrac{7.5}{1000}\cdot\dfrac{300}{0.4}\cdot8\\\\F=45\ N

So, the average recoil force on the gun during that 0.40 s burst is 45 N.

5 0
3 years ago
A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.
kow [346]

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, h_{max}, is given by the following equation;

h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}

Therefore, substituting the known values for the pebble, we have;

h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439

Therefore, the maximum height of the pebble projectile, h_{max} ≈ 6.4 m.

3 0
3 years ago
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